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Calculate the bulk modulus of air from the following data about a sound wave of wavelength 35 cm travelling in air. The pressure at a point varies between (1.0 × 10⁵ ± 14) pa and the particles of the air vibrate in simple harmonic motion of amplitude 5.5 × 10^-6 m.

Wavelength of sound wave = λ = 35 cm = 35 x 10-2 mPressure amplitude = po = (1 x 105 ± 14) paDisplacement amplitude of the air particles =So =... View Article

Find the change in the volume of 1.0 liter kerosene when it is subjected to an extra pressure of 2.0 × 10⁵ N m^-2 from the following data. Density of kerosene = 800 kg m^-3 and speed of sound in kerosene 1330 m s^-1.

The velocity in terms of the bulk modulus and density :v = √(k/p)where, k = v2 ρ=> k = (1330) 2 x 800 N/m2K = (F/A)/(Δv/Δv)Therefore, ΔV =... View Article

The absolute temperature of air in a region linearly increases from T₁ to T₂ in a space of width d. Find the time taken by a sound wave to go through the region in terms of T₁, T₂, d and the speed u of sound at 273 K. Evaluate this time for T₁ = 280 K, T₂ = 310 K, d = 33m and u = 330 m s^-1.

Temperature variation:T = T1 + [(T2 - T1)x]/dv ∝ √TvT/v = √(T/273)And, dt = dx/vT = du/v (√(273/T))We are given, initial temp = T1 = 280kFinal... View Article

At what temperature will the speed of sound be double of its value at 0°C?

T1 = 273 , v2 = 2v1, v1 = v and T2 = ?We know that, v ∝ √T=> T2/T1 = v22/v12=> T2 = 273 x 22 = 4 x 273 KSo, temp will be (4 x 273) - 273 = 819o C... View Article

The speed of sound as measured by a student in the laboratory on a winter day is 340 m s^-1 when the room temperature is 17°C. What speed will be measured by another student repeating the experiment on a day when the room temperature is 32°C?

velocity of sound = v1 = 340 m/sT1 = 17o C = 17+273 = 290 KLet v2 velocity of sound at temp T2T2 = 32o C = 32 + 273 = 305 KRelation between... View Article

Calculate the speed of sound in oxygen from the following data. The mass of 22.4 litre of oxygen at STP (T = 273 K and p = 1.0 × 10⁵ N m^-2) is 32 g, the molar heat capacity of oxygen at constant volume is Cv = 2.5 R and that at constant pressure is Cp = 3.5 R

p = 1 x 105 N/m2, T = 273 K, M = 32 g and g = 32x10-3 kgv = 22.4 l = 22.4 x10-3 m3Therefore, C/Cv = r = 3.5R/2.5R = 1.4V = √(rp/f) =... View Article

Two point sources of sound are kept at a separation of 10 cm. They vibrate in phase to produce waves of wavelength 5.0 cm. What would be the phase difference between the two waves arriving at a point 20 cm from one source

(a) on the line joining the sources and (b) on the perpendicular bisector of the line joining the sources?Solution: (a) Δx = 10 cm and λ = 5... View Article

A sound wave of frequency 100 Hz is travelling in air. The speed of sound in air is 350 m s^-1.

(a) By how much is the phase changed at a given point in 2.5 ms? (b) What is the phase difference at a given instant between two points separated... View Article

The equation of a travelling sound wave is y = 6.0 sin(600t – 1.8x) where y is measured in 10^-5 m, t in second and x in metre.

(a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave. (b) Find the ratio of the velocity amplitude of... View Article

Ultrasonic waves of frequency 4.5 MHz are used to detect tumour in soft tissues. The speed of sound in tissue is 1.5 km s^-1 and that in air is 340 m s^-1. Find the wavelength of this ultrasonic wave in air and in tissue.

Frequency of ultrasonic wave = f= 4.5 MHz or 4.5 x 106 Hz Speed of sound in tissue = 1.5 km/s Velocity of air v = 340 m/sec v= fλ λ =... View Article

Sound waves from a loudspeaker spread nearly uniformly in all directions if the wavelength of the sound is much larger than the diameter of the loudspeaker.

(a) Calculate the frequency for which the wavelength of sound in air is ten times the diameter of the speaker if the diameter is 20 cm. (b)... View Article

Find the minimum and maximum wavelengths of sound in water that is in the audible range (20—20000 Hz) for an average human ear. Speed of sound in water = 1450 m s^-1.

Solution: Speed of sound =v= 1450 m/secFor minimum wavelength, frequency should be max.f = 20 kHzWe know, v = fλor λ = 1450/[20x103] = 7.25 cmFor... View Article

A person can hear sound waves in the frequency range 20 Hz to 20 kHz. Find the minimum and the maximum wavelengths of sound that is audible to the person. The speed of sound is 360 m s^-1

Speed of sound =v= 360 m/secFrequency for minimum wavelength, f = 20 kHzWe know, v = fλor λ = 18 x 10-3 mmAgain, Frequency for max. wavelength, f... View Article

A man stands before a large wall at a distance of 50.0 m and claps his hands at regular intervals. Initially, the interval is large. He gradually reduces the interval and fixes it at a value when the echo of a clap merges with the next clap. If he has to clap 10 times during every 3 seconds, find the velocity of sound in air.

S = 50 mMan has to clap 10 times in 3 secSo, the time interval between two claps = 3/10Time taken by the sound to reach the wall = t = 3/20... View Article

At a prayer meeting, the disciples sing JAI-RAM JAI-RAM. The sound amplified by a loudspeaker comes back after reflection from a building at a distance of 80 m from the meeting. What maximum time interval can be kept between one JAI.R4M and the next JAI-RAM so that the echo does not disturb a listener sitting in the meeting. Speed of sound in air is 320 m s^-1

S = 80x2 = 160mv= 320 m/sSo, maximum time interval is: t = S/v = 160/320 = 0.5 sec... View Article

A steel tube of length 1.00 m is struck at one end. A person with his ear close to the other end hears the sound of the blow twice, one travelling through the body of the tube and the other through the air in the tube. Find the time gap between the two hearings. Use the table in the text for speeds of sounds in various substances.

Velocity of sound in air = v = 330 m/sVelocity of sound through the steel tube = vs = 5200 m/sLength of the steel tube= S = 1 mRequired time gap... View Article